Based on IR data given determine the structure of the unknown. Unknown compound A has molecular formula C4H11N. It shows a peak at 2900 cm-1 and peaks in the fingerprint region.
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Hey guys. Let's take a look at the following practice question. Here it says, based on IR data given, determine the structure of the unknown. Unknown compound A has a molecular formula of C4H11N. It shows a peak at 2,900 cm-1 and peaks in the fingerprint region. When it comes to a question like this, we refer to it as a structural determination question. But, we have to basically figure out the structure of the compound. Anytime we do a structural determination question, we must always figure out what are my degrees of unsaturation or what we call the index of hydrogen deficiency. So again, we're going to say first we find out the IHD, which is called the index of hydrogen deficiency, or your degrees of unsaturation. This will help us determine if this compound has a double bond, a triple bond, or a ring. Because when it comes to these structures, we're going to say double bond, triple bond, and then ring. When it comes to their IHD or degrees of unsaturation, we're going to say a double bond counts as 1, a triple bond counts as 2, a ring counts as 1. Now, all professors have their different formula that they use to figure out IHD. I have my own different formula as well. To figure out the IHD for a compound, we're going to use the formula 2C + H + 2 divided by 2. Remember, everyone uses a different type of IHD formula. I'm no different. Use the formula that you're most accustomed to using. Here we're going to say what do these letters represent. We're going to say C represents carbon, N represents nitrogen, C with CN with N. Now H can be more than one thing. H here can stand for hydrogen and also halogen because they both begin with H. So we're going to use this formula first to figure out what are my degrees of unsaturation or my index of hydrogen deficiency. So here, we're going to say how many carbons do we have? We have 4 times 2 is 8. We have 1 nitrogen, so that's plus 1. And we have 11 hydrogens, so that's minus 11, plus 2 divided by 2. So we're going to say what? We're going to say 8 + 1 gives me 9 but minus 11 gives me this as a negative 2. Negative 2 + 2 equals 0. What is this saying? This is saying that this unknown compound has no double bonds, no triple bonds, no ring. It is basically straight chains of carbons. What we need to realize here is IHD is 0. We have a nitrogen involved which most likely means it's an amine. Remember, we have different types of amines. If the nitrogen is connected to 1 carbon, it's primary. If the nitrogen is connected to 2 carbons, it's secondary. If it's connected to 3 carbons, we're going to say it is tertiary. It's most likely going to be one of these 3 as our answer. We just have to figure out which one it is. Now, remember, these first two, they both have NH bonds. And remember, NH bonds have their own signal. We're going to say that the signal is usually between 3,300 and 1,500 cm-1. Remember, this one would give us a double peak because there are 2 Hs on the nitrogen. This one here would give us a single peak because there's only 1 H on the nitrogen. Now, I never said anything about having a signal between 3300 and 1500 cm-1. Which means that this could not be a primary or a secondary amine, which means it has to be tertiary. What is this signal really telling me? This signal is because we have sp3 carbon-hydrogen bonds. So this signal really was useless. It doesn't tell me anything. And here, if you're in the fingerprint region, then you're definitely useless because remember, in the fingerprint regions, too many signals overlap, so we kind of avoid that region altogether, especially when it comes to structural determination. We know that it is a tertiary amine. So we're going to have the nitrogen in the middle. It's going to be connected to 3 carbons. So here's one carbon that's a methyl. Here's another carbon that's a methyl. Here's another carbon that's a methyl. Remember, there are 4 carbons involved, so make one of them an ethyl. So this would be our structure of our compound, our tertiary amine. If we wanted to name this structure, we'd say the longest carbon chain forms the end of the name. So we call this, take myself out of the image. We'd say the end of the name is ethane because it's 2 carbons. But we change the e to amine, so it's eth1-amine. The smaller alkyl groups become substituents, n substituents. So they become N,N-dimethylethanamine. Why are we saying NN? Because they're both connected to nitrogen. So that's our structure and that's the name of our structure. As you can see, structural determination questions, highly detailed. A lot goes into figuring out what it is. But you must always begin by first figuring out your index of hydrogen deficiency or your degrees of unsaturation. Same exact thing. Remember, use the formula that you are most accustomed to using to figure out what the number is, to figure out if it a double bond, a triple bond, or a ring. If it gives you an answer of 0, that means it's all single bonded carbons. None of those other three things.
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Problem
Based on IR data given determine the structure of the unknown. Unknown compound B has molecular formula C4H11N. It shows a single peak at approximately 3400 cm-1 as well as peaks at 2900 cm-1 and in the fingerprint region. Compound B also possesses a branched alkyl group.
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Hey guys. Let's take a look at the following practice question. Here it says, Based on IR data given, determine the structure of the unknown. Unknown compound B has molecular formula C4H11N. It shows a single peak at approximately 3,400 cm-1 as well as a peak at 2,900 cm-1 and in the fingerprint region. Compound B also possesses a branched alkyl group. So remember, when it comes to structural determination, what we always need to do first is to figure out the IHD, the index of hydrogen deficiency or what's called degrees of unsaturation. Now, remember, this formula, there are different formulas that all professors use. This is my version, C = 2 × c + 2. Remember, C represents carbons. N represents nitrogen. H represents hydrogens and halogens. So here we have 4 carbons, so that's 2 times 4 which is 8, plus 1 nitrogen minus 11 hydrogens. Remember here that this is going to be basically 9 minus 11, which gives us negative 2 inside the brackets. Negative 2 + 2 is 0. So this compound has no double bonds, no triple bonds, no rings. It's just a single bonded group of carbons. Now, here I tell us we have a single peak at approximately 3,400 cm-1. Remember, for a secondary amine, we say nitrogen is connected to 2 carbons and it has 1 H connected to it. Because there's 1 H connected to that nitrogen, that's going to give us a single peak. If we had a primary amine, nitrogen would be connected to only 1 carbon and because it has 2 H's on it, it would have a double peak. Here we're dealing with the secondary amine. We can draw the N in the middle. It's secondary so we know it's connected to 2 carbons and is connected to 1 H. Here I say that there's a branched group on this compound which would mean this. This would represent our branched alkyl group. So here we'd say the name for this structure if we wanted to name it, we'd say this is 1-2-3. That's the larger alkyl chain. This is the smaller alkyl chain so the whole thing becomes a substituent. So here we're going to say that the name of this structure is going to be what? It's going to be, take myself out of the image, it's going to be a propane chain. Remember, it's an amine, so you change the 'e' from propane to amine, so it's propanamine. We're going to say the smaller alkyl group is called an N-substituent, so it'll be N-methyl. The nitrogen is on carbon number 2, so it's actually going to be N-methyl-2-propanamine. Again, it's a single peak, so it's a secondary amine. We figured out the IHD to 0, so there's no double bond, no triple bond, no ring. Branched alkyl changes mean here that we have a branch group which is really this coming off of that chain which lengthens the chain, the 3 carbons. This would be our structure. Again, structural determination when we move over to NMR is difficult. Even IR structural determination is difficult. But you have to remember the steps to take. Always find IHD first. Since we're dealing with IR here, remember, when they give me a signal, which functional group are they alluding to? Knowing that is the key to figuring out what the compound looks like.
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Problem
Based on IR data given determine the structure of the unknown. Unknown compound C has molecular formula C6H10O3. It shows peaks at 2900, 1850 , 1740 cm-1 and in the fingerprint region.
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Hey guys. Let's take a look at the following question. Here it says based on IR data given, Unknown compound C has molecular formula C6H10O3. It shows peaks at 2900, 1850, and 1740 centimeters-1 and in the fingerprint region. So remember, anytime we're given a structural determination question, we must always first begin by figuring out the IHD, also called your degrees of unsaturation. Remember, all of us use different formulas. My particular formula is C+NH+H+22. C represents carbon, N represents nitrogen, and H represents hydrogens and halogens. So we have 6 carbons here. So it's 6 times 2 which gives me 12. There are no hydrogens, in brackets, plus 2 divided by 2. Oxygen is not part of my formula. Therefore, I ignore it as well. So here we're going to say 12 minus 10 gives me 2, plus 2 divided by 2, that's 4 divided by 2 which gives me an IHD of 2. Now, here we say we have 1740 which means that we have a carbonyl. But more specifically, it means we have an ester because an ester falls within this region. Because an ester is 1730-1750 centimeters-1. Next, we're going to say this 2900 really isn't telling me much because this represents sp3 carbon-hydrogen bonds. So here we're going to say that signal there is kind of useless. Now, here we have 1850 signal which also really doesn't tell me much at all. So, what we're going to have to do is try to think of the possible shapes that this compound could take. Now, when we get to NMR, we'll be more specific in terms of what the structure could be. But because I only gave you IR information, this opens up the possibility of having multiple answers for this particular question. IHD of 2. We know that one of the IHD comes from the fact that we have a carbonyl of the ester. We could say here, we have 3 oxygens so that carbonyl has both oxygens. Remember, to be an ester, those oxygens on the end have to be connected to carbons. Now, I can't do straight chains for them because here it wouldn't give me an IHD of 2. So that second IHD that I'm getting, I can assume maybe it's coming from a ring. This takes care of 4 of my carbons. But how many carbons do we have? We have 6 carbons, so I need to draw 2 more. So we could say potentially, maybe we have 2 methyl groups on one of these carbons. This molecule has a lot of symmetry, so this could be a possible answer. But remember, I'm only giving us IR, so it opens up the possibility to multiple answers. Maybe both those methyl groups are not on the same carbon. Maybe one is over here and one is over here. The molecule still has symmetry, but it's different. When we get to NMR, we'll be given more information, more signals, which will narrow it down to one particular answer. For a question like this, we're basically, in a sense, guessing which one is the best structure based on the information that I've provided you. Because I make it so vague, more than one possible answer exists.
4
Problem
Match the following functional group choices with the supplied infrared spectra data
A
Ether
B
Ketone
C
Alcohol
D
Alkene
E
Nitrile
5
Problem
Match the following functional group choices with the supplied infrared spectra data
A
Alkyl Halide
B
Alkyne
C
Carboxylic Acid
D
Alkene
E
Ketone
6
Problem
Match the following functional group choices with the supplied infrared spectra data.
A
Aldehyde
B
Alkane
C
Carboxylic acid
D
Ester
E
Ether
7
Problem
Match the following functional group choices with the supplied infrared spectra data.
